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Chapter_5

# Chapter_5 - 5-19C Energy can be transferred to or from a...

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5-19C Energy can be transferred to or from a control volume as heat, various forms of work, and by mass. 5-24E A water pump increases water pressure. The flow work required by the pump is to be determined. Assumptions 1 Flow through the pump is steady. 2 The state of water at the pump inlet is saturated liquid. 3 The specific volume remains constant. Properties The specific volume of saturated liquid water at 10 psia is /lbm ft 01659 . 0 3 psia 10 @ f v v (Table A-5E) Then the flow work relation gives Btu/lbm 0.1228 3 3 1 2 1 1 2 2 flow ft psia 5.404 Btu 1 10)psia /lbm)(50 ft 01659 . 0 ( ) ( P P P P w v v v 5-29C The kinetic energy of a fluid increases at the expense of the internal energy as evidenced by a decrease in the fluid temperature. 5-32 EES Problem 5-31 is reconsidered. The effect of the inlet area on the mass flow rate, exit velocity, and the exit area as the inlet area varies from 50 cm 2 to 150 cm 2 is to be investigated, and the final results are to be plotted against the inlet area. Analysis The problem is solved using EES, and the solution is given below. Function HCal(WorkFluid\$, Tx, Px) "Function to calculate the enthalpy of an ideal gas or real gas" If 'Air' = WorkFluid\$ then HCal:=ENTHALPY('Air',T=Tx) "Ideal gas equ." else HCal:=ENTHALPY(WorkFluid\$,T=Tx, P=Px) "Real gas equ." endif end HCal "System: control volume for the nozzle" "Property relation: Air is an ideal gas" "Process: Steady state, steady flow, adiabatic, no work" "Knowns - obtain from the input diagram" WorkFluid\$ = 'Air' T[1] = 200 [C] P[1] = 300 [kPa] Vel[1] = 30 [m/s] P[2] = 100 [kPa] Vel[2] = 180 [m/s] A[1]=80 [cm^2] Am[1]=A[1]*convert(cm^2,m^2) "Property Data - since the Enthalpy function has different parameters for ideal gas and real fluids, a function was used to determine h." h[1]=HCal(WorkFluid\$,T[1],P[1]) h[2]=HCal(WorkFluid\$,T[2],P[2]) "The Volume function has the same form for an ideal gas as for a real fluid." v[1]=volume(workFluid\$,T=T[1],p=P[1]) v[2]=volume(WorkFluid\$,T=T[2],p=P[2]) Water 10 psia 50 psia

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"Conservation of mass: " m_dot[1]= m_dot[2] "Mass flow rate" m_dot[1]=Am[1]*Vel[1]/v[1] m_dot[2]= Am[2]*Vel[2]/v[2] "Conservation of Energy - SSSF energy balance" h[1]+Vel[1]^2/(2*1000) = h[2]+Vel[2]^2/(2*1000) "Definition" A_ratio=A[1]/A[2] A[2]=Am[2]*convert(m^2,cm^2) A 1 [cm 2 ] A 2 [cm 2 ] m 1 T 2 50 24.19 0.3314 184.6 60 29.02 0.3976 184.6 70 33.86 0.4639 184.6 80 38.7 0.5302 184.6 90 43.53 0.5964 184.6 100 48.37 0.6627 184.6 110 53.21 0.729 184.6 120 58.04 0.7952 184.6 130 62.88 0.8615 184.6 140 67.72 0.9278 184.6 150 72.56 0.9941 184.6 50 70 90 110 130 150 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 A[1] [cm^2] m [1]
5-38E Air is decelerated in a diffuser from 600 ft/s to a low velocity. The exit temperature and the exit velocity of air are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with variable specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heat transfer is negligible. 5 There are no work interactions.

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